Question

Ao calcular a integral iterada 2016.2-U1S2-AAP-CDI3-Q10.jpg, chega-se ao seguinte resultado.

Answer 1

Boa tarde!

Solução!

No final x vai ficar em função de z.


$\displaystyle\int _{0}^{2}\displaystyle\int_{0}^{z}\int_{0}^{x+z} 2ydzdydx\\\\\\\\\\ \displaystyle\int _{0}^{2}\displaystyle dx \displaystyle\int_{0}^{z}2ydy\bigg[x+z\bigg] _{0} ^{x+z} \\\\\\\\\\\\ \displaystyle\int _{0}^{2}\displaystyle dx 2\displaystyle\int_{0}^{z}(yx+yz)dy\\\\\\\\\\\\ 2\displaystyle\int _{0}^{2}dx\bigg[yx+yz\bigg]_{0}^{z}\\\\\\\\\\\ 2\displaystyle\int _{0}^{2}dx\bigg[ \frac{y^{2} }{2} x+ \frac{ y^{2} }{2} z\bigg]_{0}^{z}$


$2\displaystyle\int _{0}^{2}dx\bigg[ \frac{z^{2} }{2} x+ \frac{ z^{2} }{2} z\bigg]_{0}^{z}\\\\\\\ 2\displaystyle\int _{0}^{2}dx\bigg[ \frac{z^{2} }{2} x+ \frac{ z^{3} }{2} \bigg]\\\\\\\ 2. \dfrac{ z^{2} }{2} \bigg[x+z\bigg]_{0}^{2}\\\\\\\\ 2. \dfrac{ z^{2} }{2} \bigg[ \frac{ x^{2} }{2} +z\bigg]_{0}^{2}\\\\\\\\ 2. \dfrac{ z^{2} }{2} \bigg[ \frac{ 2^{2} }{2} +z\bigg]\\\\\\\\ 2. \dfrac{ z^{2} }{2} \bigg[ \frac{ 4 }{2} +z\bigg]$


$2. \dfrac{ z^{2} }{2} \bigg[2 +z\bigg]\\\\\\\\ 2z^{2}\bigg[1+z\bigg]\\\\\\\\ 2(2)^{2}\bigg[1+2\bigg]\\\\\\\\ 2.(4)\bigg[3\bigg]\\\\\\\\ 8.3=24$


$\boxed{Resposta:~~\displaystyle\int _{0}^{2}\displaystyle\int_{0}^{z}\int_{0}^{x+z} 2ydzdydx~~=24~~\boxed{Alternativa~~C}}$

Boa noite!
Bons estudos!